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8.9: Absorption and Stimulated Emission of Radiation

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Let us use some of the results of time-dependent perturbation theory to investigate the interaction of an atomic electron with classical (i.e., non-quantized) electromagnetic radiation.

The unperturbed Hamiltonian is

$ H_0 = \frac{p^2}{2 \,m_e} + V_0(r).$ \ref{862}

The standard classical prescription for obtaining the Hamiltonian of a particle of charge $ q$ in the presence of an electromagnetic field is

$ {\bf p}$ $ \rightarrow {\bf p} - q\,{ \bf A},$ \ref{863} $ H$ $ \rightarrow H-q\,\phi,$ \ref{864}

where $ {\bf A}(\bf r)$ is the vector potential and $ \phi({\bf r})$ is the scalar potential. Note that

$ {\bf E}$ $ = - \nabla\phi - \frac{\partial {\bf A}}{\partial t},$ \ref{865} $ {\bf B}$ $ = \nabla\times {\bf A}.$ \ref{866}

This prescription also works in quantum mechanics. Thus, the Hamiltonian of an atomic electron placed in an electromagnetic field is

$ H = \frac{\left\vert{\bf p} + e\, {\bf A}\right\vert^{\,2} }{2\,m_e}- e \,\phi + V_0(r),$ \ref{867}

where $ {\bf A}$ and $ \phi$ are real functions of the position operators. The above equation can be written

$ H = \frac{ \left(p^2 +e \,{\bf A}\cdot {\bf p} +e \,{\bf p}\cdot{\bf A} + e^2 A^2\right)}{2\,m_e}- e \,\phi + V_0(r).$ \ref{868}

Now,

$ {\bf p}\cdot {\bf A} = {\bf A}\cdot {\bf p},$ \ref{869}

provided that we adopt the gauge $ \nabla\cdot{\bf A} = 0$ . Hence,

$ H = \frac{p^2}{2\,m_e} -\frac{e\,{\bf A}\cdot{\bf p}}{m_e} +\frac{ e^2 A^2}{2\,m_e}- e\, \phi + V_0(r).$ \ref{870}

Suppose that the perturbation corresponds to a monochromatic plane-wave, for which

$ \phi$ $ = 0,$ \ref{871} $ {\bf A}$ $ = 2\, A_0 \,$$ \mbox{\boldmath$$ \,\cos\left (\frac{\omega}{c} \,{\bf n}\cdot{\bf x} - \omega\, t\right),$ \ref{872}

where $ \epsilon$ and $ {\bf n}$ are unit vectors that specify the direction of polarization and the direction of propagation, respectively. Note that $ \epsilon$ $ \cdot{\bf n} = 0$ . The Hamiltonian becomes

$ H = H_0 + H_1(t),$ \ref{873}

with

$ H_0 = \frac{p^2}{2\,m_e} + V(r),$ \ref{874}

and

$ H_1 \simeq \frac{e\,{\bf A}\cdot{\bf p}}{m_e},$ \ref{875}

where the $ A^2$ term, which is second order in $ A_0$ , has been neglected.

The perturbing Hamiltonian can be written

$H_1 = \frac{e \,A_0\, \mbox{\boldmath$\epsilon$}\cdot{\bf p} }{m_... ...\exp[-{\rm i}\,(\omega/c)\, {\bf n}\cdot{\bf x} + {\rm i}\, \omega\, t]\right).$ \ref{876}

This has the same form as Equation \ref{850}, provided that

$ V = - \frac{e \,A_0\, \mbox{\boldmath$ \ref{877}

It is clear, by analogy with the previous analysis, that the first term on the right-hand side of Equation \ref{876} describes the absorption of a photon of energy $ \hbar\,\omega$ , whereas the second term describes the stimulated emission of a photon of energy $ \hbar\,\omega$ . It follows from Equations \ref{859} and \ref{860} that the rates of absorption and stimulated emission are

$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \frac{e^2}{m_e^{\,2}}\, \... ...\,2}\, \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i -\hbar\,\omega),$ \ref{878}

and

$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \frac{e^2}{m_e^{\,2}}\, \... ...{\,2}\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i +\hbar\,\omega),$ \ref{879}

respectively.

Now, the energy density of a radiation field is

$ U = \frac{1}{2}\left(\frac{\epsilon_0\,E_0^{\,2}}{2}+ \frac{B_0^{\,2}}{2\,\mu_0} \right),$ \ref{880}

where $ E_0$ and $ B_0=E_0/c= 2\,A_0\,\omega/c$ are the peak electric and magnetic field-strengths, respectively. Hence,

$ U = 2\,\epsilon_0\,c\,\omega^2\,\vert A_0\vert^{\,2},$ \ref{881}

and expressions \ref{878} and \ref{879} become

$w_{i\rightarrow n} = \frac{\pi}{\hbar} \frac{e^2}{\epsilon_0\,m_e... ...\, U\, \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i -\hbar\,\omega),$ \ref{882}

and

$w_{i\rightarrow n} = \frac{\pi}{\hbar} \frac{e^2}{\epsilon_0\,m_e... ...}\, U\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}\, \delta(E_n-E_i +\hbar\,\omega),$ \ref{883}

respectively. Finally, if we imagine that the incident radiation has a range of different frequencies, so that

$ U = \int d\omega\,u(\omega),$ \ref{884}

where $ d\omega\,u(\omega)$ is the energy density of radiation whose frequency lies in the range $ \omega$ to $ \omega+d\omega$ , then we can integrate our transition rates over $ \omega$ to give

$w_{i\rightarrow n} = \frac{\pi}{\hbar^2} \frac{e^2}{\epsilon_0\,m... ..., \vert\langle n\vert\, \exp[\,{\rm i}\,(\omega_{ni}/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}$ \ref{885}

for absorption, and

$w_{i\rightarrow n} = \frac{\pi}{\hbar^2} \frac{e^2}{\epsilon_0\,m... ...\, \vert\langle n\vert\, \exp[-{\rm i}\,(\omega_{in}/c)\,{\bf n}\cdot{\bf x}]\,$ $ \mbox{\boldmath$$ \cdot{\bf p} \,\vert i\rangle\vert^{\,2}$ \ref{886}

for stimulated emission. Here, $ \omega_{ni} = (E_n-E_i)/\hbar>0$ and $ \omega_{in} = (E_i-E_n)/\hbar>0$ . Furthermore, we are assuming that the radiation is incoherent, so that intensities can be added.

Contributors

Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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